Answer to Question #343366 in Statistics and Probability for Sow

Question

manager of a courier service believes that packets delivered at the end of the month are heavier than those delivered early in the month. As an experiment, he weighed a random sample of 20 packets at the beginning of the month. He found that the mean weight was 5.25 kgs with a standard deviation of 1.20 kgs. Ten packets randomly selected at the end of the month had a mean weight of 4.96 kgs and a standard deviation of 1.15 kgs. At the 0.05 significance level, can it be concluded that the packets delivered at the end of the month weigh more?

Answer 1

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{1.20^2}{1.15^2}=1.0888$$

The critical values for $\alpha=0.05,$ $df_1=n_1-1=19$ degrees of freedom, $df_2=n_2-1=9$ degrees of freedom are $F_L = 0.3472$ and $F_U = 3.6833,$ and since $F = 1.0888$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\ge\mu_2$

$H_1:\mu_1<\mu_2$

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

The degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=df_1+df_2=n_1+n_2-2=28$$

Based on the information provided, the significance level is $\alpha = 0.05,$ and the degrees of freedom are $df = 28.$

The critical value for this left-tailed test, $\alpha = 0.05, df=153$ degrees of freedom is $t_c =-1.701131.$

The rejection region for this left-tailed test is $R = \{t: t<- 1.701131\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{5.25-4.96}{\sqrt{\dfrac{(20-1)(1.20)^2+(10-1)(1.15)^2}{20+10-2}(\dfrac{1}{20}+\dfrac{1}{10})}}$$

$$=0.6323$$

Since it is observed that $t =0.6323>-1.701131= t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for left-tailed, $df=28$ degrees of freedom, $t=0.6323$ is $p= 0.73384,$ and since $p= 0.73384>0.05=\alpha,$ it is concluded that the null hypothes is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is less than $\mu_2,$ at the $\alpha = 0.05$ significance level.